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Integral Calculus

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Joined:	19 Feb 2007
Post:	3834

19 Mar 2007 15:29:26 IST

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644

area again!!!!!!!!!!!

None

find the area by curve

y = sin²x (also sketch it)

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Kiran H

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Joined: 7 Nov 2006

Posts: 296

19 Mar 2007 16:09:24 IST

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write (sin x)^2 as (1 - cos 2x )/ 2 and then integrate.

(1 - COS(2·x) ) /2 dx

= x/2 - SIN(2·x) /4
Now substitute the limits and find the area. If the limits are 0 to pi then area is

A= pi / 2 .

Cheers!

Himanshu

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Joined: 19 Feb 2007

Posts: 3834

19 Mar 2007 16:13:47 IST

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thanx kiran!!!

kannan kv

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Joined: 13 Mar 2007

Posts: 130

19 Mar 2007 16:26:01 IST

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hey there is a simpler method (reduction formula)

_[0]

^[pi/2] sinⁿ x dx = (n-1/n).(n-3/n-2).(n-5/n-4)....2/3        if n is odd

                          = (n-1/n).(n-3/n-2).(n-5/n-4).....(1/2).pi/2   if n is even

eg.

_[0]

^[pi/2] sin⁵ x dx = (4/5).(2/3) (this it sooo simple isnt it)

for sin² x

_[0]

^[pi/2] sin² x dx=(1/2).(pi/2)

note: this method is applicable only for 0 to pi/2 .for other limits like 0 to pi or 0 to pi/4 use substitution method.
eg
for 0 to pi take t=x/2 & proceed

for 0 to pi/4 take t=2x & proceed

for 0 to pi/6 take t=3x & proceed.....

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